Lenses: Types, Ray Diagrams, Lens Formula & Applications

What is a Lens?

A lens is a transparent piece of glass or plastic with at least one curved surface. It bends (refracts) light rays to form images. Lenses are everywhere: in glasses, cameras, microscopes, and telescopes.

Types of Lenses

There are two main types of lenses:

1. Convex Lens (Converging Lens)

   • Thicker in the middle, thinner at the edges.
   • Bends parallel rays inward to meet at a point (focus).
   • Used in magnifying glasses, cameras, and projectors.

2. Concave Lens (Diverging Lens)

   • Thinner in the middle, thicker at the edges.
   • Bends parallel rays outward, making them appear to diverge from a point (virtual focus).
   • Used in spectacles for short-sightedness and in optical instruments.

Key Terms and Definitions

1. Optical Centre (O)

• The central point of the lens.
• A ray passing through O goes straight without bending.

2. Principal Axis

• An imaginary straight line passing through O and the centres of curvature of the lens surfaces.

3. Focus (F) and Focal Length (f)

• Convex Lens: Parallel rays meet at the real focus (F) after refraction. The distance from O to F is the focal length (f).
• Concave Lens: Parallel rays appear to diverge from the virtual focus (F). The focal length is negative by convention.

4. Object Distance (u)

• Distance from the object to the lens (always negative for real objects in lens formula conventions).

5. Image Distance (v)

• Distance from the lens to the image.
• Positive for real images (formed on the opposite side of the object).
• Negative for virtual images (formed on the same side as the object).

6. Magnification (m)

• Ratio of image height to object height.
• Formula: $m=\frac{v}{u}=\frac{\text{Image Height}}{\text{Object Height}}$.
• Positive m: Image is upright.
• Negative m: Image is inverted.

Rules for Ray Tracing

To locate the image formed by a lens, use these three rules for ray tracing:

1. Ray Parallel to Principal Axis

   • After refraction, it passes through the focus (F) for a convex lens.
   • For a concave lens, it appears to diverge from the virtual focus (F).

2. Ray Passing Through the Optical Centre (O)

   • Goes straight without bending.

3. Ray Passing Through the Focus (for Convex Lens) or Directed Towards the Focus (for Concave Lens)

   • After refraction, it emerges parallel to the principal axis.

flowchart TD
    A["Ray 1: Parallel to Principal Axis"] -->|"Convex Lens"| B["Passes through F"]
    A -->|"Concave Lens"| C["Appears to diverge from F"]
    D["Ray 2: Through O"] --> E["Goes straight"]
    F["Ray 3: Through F (Convex) or Towards F (Concave)"] -->|"Convex"| G["Parallel to Principal Axis"]
    F -->|"Concave"| H["Parallel to Principal Axis"]

Lens Formula and Magnification

The lens formula relates object distance (u), image distance (v), and focal length (f):

$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$

Magnification (m) is given by: $$m=\frac{v}{u}$$

Sign Conventions

Solved Examples

Example 1: Convex Lens

Problem: An object is placed 20 cm from a convex lens of focal length 10 cm. Find the image distance and magnification.

Solution: Given:

• $u=-20$ cm (object is real),
• $f=+10$ cm.

Using the lens formula: $$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \frac{1}{10}=\frac{1}{v}-\frac{1}{-20} \\ \frac{1}{10}=\frac{1}{v}+\frac{1}{20} \\ \frac{1}{v}=\frac{1}{10}-\frac{1}{20}=\frac{2-1}{20}=\frac{1}{20} \\ v=20\text{ cm (positive, so real image)} \end{gathered}$$

Magnification: $$m=\frac{v}{u}=\frac{20}{-20}=-1$$

• Image distance (v): 20 cm (real, inverted, same size as object).

Example 2: Concave Lens

Problem: An object is placed 15 cm from a concave lens of focal length 10 cm. Find the image distance and magnification.

Solution: Given:

• $u=-15$ cm,
• $f=-10$ cm.

Using the lens formula: $$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \frac{1}{-10}=\frac{1}{v}-\frac{1}{-15} \\ -\frac{1}{10}=\frac{1}{v}+\frac{1}{15} \\ \frac{1}{v}=-\frac{1}{10}-\frac{1}{15}=-\frac{3-2}{30}=-\frac{1}{30} \\ v=-30\text{ cm (negative, so virtual image)} \end{gathered}$$

Magnification: $$m=\frac{v}{u}=\frac{-30}{-15}=+2$$

• Image distance (v): 30 cm (virtual, upright, magnified).

Comparison of Convex and Concave Lenses

Applications of Lenses

1. Convex Lenses:

   • Magnifying Glass: Produces a magnified virtual image.
   • Camera: Forms a real, inverted image on the film/sensor.
   • Projector: Produces a magnified real image on a screen.
   • Spectacles: Corrects long-sightedness (hypermetropia).

2. Concave Lenses:

   • Spectacles: Corrects short-sightedness (myopia).
   • Optical Instruments: Used in combination with convex lenses to reduce aberrations.

mindmap
  root((Lenses))
    Convex Lens
      Magnifying Glass
      Camera
      Projector
      Spectacles (Hypermetropia)
    Concave Lens
      Spectacles (Myopia)
      Optical Instruments

NEB Board-Style Questions

Short Answer Questions

1. Define focal length of a lens. How does it differ for convex and concave lenses?

   • Answer: Focal length is the distance between the optical centre and the focus. For convex lenses, it is positive; for concave lenses, it is negative.

2. State the lens formula. What does a negative magnification indicate?

   • Answer: Lens formula: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$. Negative magnification indicates the image is inverted.

3. Why is a concave lens used in spectacles for short-sightedness?

   • Answer: A concave lens diverges light rays, shifting the image back to the retina, which is too far forward in short-sighted eyes.

Long Answer Questions

4. An object of height 5 cm is placed 10 cm in front of a convex lens of focal length 15 cm. Find the position, nature, and size of the image formed.

   • Solution: Given: $u=-10$ cm, $f=15$ cm, object height $h_o=5$ cm. $$\begin{gathered} \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{15}+\frac{1}{-10}=\frac{2-3}{30}=-\frac{1}{30} \\ v=-30\text{ cm (virtual image)} \end{gathered}$$ Magnification: $$\begin{gathered} m=\frac{v}{u}=\frac{-30}{-10}=3 \\ \text{Image height}=m\times h_o=3\times 5=15\text{ cm (upright, magnified)} \end{gathered}$$
   • Answer: The image is 30 cm from the lens on the same side as the object, virtual, upright, and 15 cm tall.

5. Draw ray diagrams to show the formation of a virtual image by a concave lens when the object is placed in front of it.

   • Answer: Use the three ray rules for a concave lens (as shown in the earlier figure).

Exam Tip

1. Memorize the Lens Formula: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ and magnification $m=\frac{v}{u}$.
2. Sign Conventions: Always use the correct signs for $u$, $v$, and $f$ (negative for real objects and concave lenses).
3. Ray Diagrams: Practice drawing ray diagrams for both convex and concave lenses to visualize image formation.
4. Applications: Know the practical uses of lenses in daily life (e.g., spectacles, cameras).
5. Numerical Problems: Solve problems step-by-step, showing all calculations clearly. Watch out for units (always use cm) and signs.

A standard labelled diagram of a convex lens showing rays, focus, and image formation. (Image: Chetvorno, CC0, via Wikimedia Commons) A standard labelled diagram of a concave lens showing rays, virtual focus, and image formation. (Image: Takuzaburou, Public domain, via Wikimedia Commons)